When two tuning forks were sounded together , 20 beats were produced in 10 seconds. On loading one of the fork with wax, the number of beats increases. If the frequency of unloaded fork is 512Hz , calculate the frequency of other.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the initial beat frequency
When two tuning forks are sounded together, the number of beats produced is given. In this case, 20 beats were produced in 10 seconds.

**Calculation:**
\[ \text{Beat frequency} = \frac{\text{Number of beats}}{\text{Time in seconds}} = \frac{20 \text{ beats}}{10 \text{ seconds}} = 2 \text{ beats per second} \]

### Step 2: Understand the relationship between frequencies and beats
The beat frequency is the absolute difference between the frequencies of the two tuning forks. Let’s denote the frequency of the unloaded fork as \( f_B = 512 \text{ Hz} \) and the frequency of the other fork as \( f_A \). The beat frequency can be expressed as:
\[ |f_A - f_B| = 2 \text{ Hz} \]
This means:
\[ f_A - 512 = 2 \quad \text{or} \quad 512 - f_A = 2 \]

### Step 3: Solve for \( f_A \)
From the first equation:
\[ f_A - 512 = 2 \implies f_A = 512 + 2 = 514 \text{ Hz} \]

From the second equation:
\[ 512 - f_A = 2 \implies f_A = 512 - 2 = 510 \text{ Hz} \]

### Step 4: Analyze the effect of loading one fork with wax
When one of the forks is loaded with wax, its frequency decreases. Since the problem states that the number of beats increases after loading one fork, it implies that the frequency of the loaded fork (let's say \( f_A \)) must be less than the frequency of the unloaded fork \( f_B \).

Thus, the frequency of the other fork \( f_A \) must be:
\[ f_A = 510 \text{ Hz} \]

### Conclusion
The frequency of the other tuning fork is \( 510 \text{ Hz} \).

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